• Title: Integrable Functions

• Series: Fourier Transform

• Chapter: Fourier Series

• YouTube-Title: Fourier Transform 5 | Integrable Functions

• Bright video: https://youtu.be/2NUNRQEp7YE

• Dark video: https://youtu.be/LgRBO95N0xM

• Subtitle on GitHub: ft05_sub_eng.srt missing

• Timestamps (n/a)
• Subtitle in English (n/a)
• Quiz Content

Q1: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{x}$ for $x > 0$. Is the function integrable?

A1: No, because $\int_0^1 | f(x) | , dx = \infty$.

A2: Yes, because $\int_0^1 | f(x) | , dx < \infty$.

A3: No, because $f$ is unbounded.

A4: Yes, because $f$ is bounded.

Q2: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function integrable?

A1: No, because $\int_0^1 | f(x) | , dx = \infty$.

A2: Yes, because $\int_0^1 | f(x) | , dx < \infty$.

A3: No, because $f$ is unbounded.

A4: Yes, because $f$ is bounded.

Q3: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function square-integrable?

A1: No, because $\int_0^1 | f(x) |^2 , dx = \infty$.

A2: Yes, because $\int_0^1 | f(x) |^2 , dx < \infty$.

A3: No, because $f$ is unbounded.

A4: Yes, because $f$ is bounded.

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